The area bounded by the curve y^2 = 12x and the line x=3 is revolved about the line x=3. What is the volume generated?
Hi, can anyone do it it disk method in terms of dx plss
Well, I would say that revolving around the line x=3 sounds like a real spin-off. But let's not go off on a tangent here, we need to find the volume generated.
To find the volume, we can use the method of cylindrical shells. If we revolve the region around the line x=3, we create a cylindrical shell for every point on the curve.
First, let's express the curve equation in terms of y: y = ±√(12x). Now, we want to find the bounds of integration for y. Since we're revolving around x=3, the bounds will be from y=-√(12*3) to y=√(12*3).
Next, we can express the curve equation in terms of x: x = y^2/12. So, the distance from the axis of rotation to the curve will be x = 3 - y^2/12.
Now, it's time to calculate the volume. The volume of a cylindrical shell is given by V = 2πrhΔx, where r represents the distance to the axis of rotation, h represents the height of the cylindrical shell, and Δx represents the width of the shell.
Integrating with respect to y, we get: V = 2π ∫[from -√(12*3) to √(12*3)] [(3 - y^2/12) * y] dy.
After evaluating this integral, you should get the volume generated by revolving around x=3. So, get busy and solve it, and remember to enjoy the process - it's the curvature that counts!
To find the volume generated, we can use the method of cylindrical shells.
Step 1: Determine the limits of integration.
Since we are rotating the area bounded by the curve y^2 = 12x and the line x=3 about the line x=3, we need to find the range of y-values. From the given equation, we have y^2 = 12x. Simplifying and solving for y, we get y = ±√(12x). Since the curve is symmetric about the x-axis, we only need to consider the positive values of y. So, the limits of integration for y are 0 to √(12x).
Step 2: Set up the integral.
To find the volume, we need to integrate the circumference of each shell multiplied by its height. The circumference of a shell at a given x-value is 2πr, where r is the distance from the shell to the axis of rotation (x=3). In this case, r = x - 3. The height of each shell is given by √(12x) - 0, or simply √(12x).
Therefore, the integral to find the volume is:
V = ∫[from x=0 to x=9] (2π(x - 3) * √(12x)) dx
Step 3: Evaluate the integral.
Integrating the expression, we get:
V = ∫[from x=0 to x=9] (2π(x - 3) * √(12x)) dx
= 2π * ∫[from x=0 to x=9] (x√(12x) - 3√(12x)) dx
To simplify the integration, we can split it into two parts:
V = 2π * (∫[from x=0 to x=9] (x√(12x)) dx - ∫[from x=0 to x=9] (3√(12x)) dx)
We can solve each integral separately using the power rule for integration:
∫(x^n) dx = (1/(n+1)) * x^(n+1)
Applying the power rule, we have:
V = 2π * ((1/4) * ∫[from x=0 to x=9] (12x^(3/2)) dx - (1/4) * 3 * ∫[from x=0 to x=9] (12x^(1/2)) dx)
= π/2 * (∫[from x=0 to x=9] (3x^(3/2)) dx - 9 * ∫[from x=0 to x=9] (x^(1/2)) dx)
Evaluating the integrals:
V = π/2 * [(2/5) * (3x^(5/2)) | [from x=0 to x=9] - 9 * (2/3) * (x^(3/2)) | [from x=0 to x=9)]
= π/2 * [(2/5) * 3 * (9^(5/2) - 0) - 9 * (2/3) * (9^(3/2) - 0)]
= π/2 * [(2/5) * 3 * (27√3) - 9 * (2/3) * (27)]
= π/2 * [54√3 - 54]
= (27√3 - 27)π
So, the volume generated when the area bounded by the curve y^2 = 12x and the line x=3 is revolved about the line x=3 is (27√3 - 27)π.
To find the volume generated by revolving the area bounded by the curve y^2 = 12x and the line x=3 about the line x=3, we will use the method of cylindrical shells.
First, let's sketch the graph of the curve y^2 = 12x:
y^2 = 12x can be rewritten as y = √(12x) or y = -√(12x).
We can plot these two functions on a graph and observe that they are symmetrical with respect to the x-axis and intersect at (0,0) and (12,±4√3).
Now we need to find the height and the radius of the cylindrical shells.
The height of each cylindrical shell is the difference between the upper and lower curves. In this case, the upper curve y = √(12x) and the lower curve y = -√(12x). Therefore, the height is 2√(12x).
The radius of each cylindrical shell is the distance from the axis of rotation to the curve, which in this case is x = 3.
Now, we integrate the expression for the circumference of a cylindrical shell multiplied by its height over the interval where the curve intersects the line x = 3.
The interval is from x = 0 to x = 12 since the upper curve intersects x = 3 at x = 12.
The integral for the volume generated is given by:
V = ∫[0, 12] (2π(3)(2√(12x))) dx
Simplifying further:
V = ∫[0, 12] 12π√(12x) dx
V = 12π ∫[0, 12] √(12x) dx
To evaluate this integral, we can use a substitution.
Let u = 12x, then du = 12 dx.
The integral becomes:
V = 12π ∫[0, 12] √(u) (1/12) du
V = π ∫[0, 12] √(u) du
Integrating further:
V = π * (2/3) u^(3/2) | [0, 12]
V = π * (2/3) (12)^(3/2) - π * (2/3) (0)^(3/2)
V = π * (2/3) (6^3) - π * (2/3) * 0
V = π * (2/3) (216)
V = π * (432/3)
V = 144π
Therefore, the volume generated by revolving the area bounded by the curve y^2 = 12x and the line x=3 about the line x=3 is 144π cubic units.
x = 3 and y^2 = 12x intersect at (3,6) and (3,-6)
so we can take horizontal slices, and because of the symmetry just go from y = 0 to y = 6 and double the result.
we also have a radius of 3 - y^2/12
V = 2π ∫r^2 dy from 0 to 6
= 2π ∫(3 - y^2/12)^2 dy from 0 to 6
= 2π ∫(9 - (1/2)y^2 + y^4/144) dy
= 2π [ 9y - (1/6)y^3 + (1/720)y^5] from 0 to 6
= 2π ( 54 - 216/6 + 7776/720 - 0 )
= 2π(144/5)
= 288π/5 or appr 180.96
confirmation:
http://www.wolframalpha.com/input/?i=integral+2%CF%80%283+-+x%5E2%2F12%29%5E2++from+0+to+6